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Zorluk: OrtaMeasures of Dispersion

A laboratory technician recorded the temperature changes (in °C) of a chemical reaction across five trials as 33, 66, 77, 99, and 1515. What is the variance of this set of data?

Cevap: 16

Cevap

The variance of the temperature changes is 16.
The mean of the given numbers is xˉ=3+6+7+9+155=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8. The squared deviations from the mean are (3−8)2=25(3-8)^2 = 25, (6−8)2=4(6-8)^2 = 4, (7−8)2=1(7-8)^2 = 1, (9−8)2=1(9-8)^2 = 1, and (15−8)2=49(15-8)^2 = 49. Summing these squared deviations gives 25+4+1+1+49=8025 + 4 + 1 + 1 + 49 = 80. Dividing by the number of observations N=5N = 5 yields the variance: 805=16\frac{80}{5} = 16.

Adım Adım Çözüm

1
Calculate the arithmetic mean (\bar{x}) of the dataset.
\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8
The mean is required to determine the deviation of each individual value from the central value.
2
Compute the square of the deviation for each data point from the mean.
(3-8)^2 = 25, (6-8)^2 = 4, (7-8)^2 = 1, (9-8)^2 = 1, (15-8)^2 = 49
Squaring deviations ensures all values are positive and emphasizes larger departures from the mean.
3
Sum the squared deviations and divide by the total number of observations (N = 5).
\text{Variance} = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16
Variance measures the average of the squared deviations from the mean.

Anahtar Kavram

Population Variance for Ungrouped Data
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Konular

Measures of Dispersion

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